How do you simplify #7sqrt32#? Algebra Radicals and Geometry Connections Simplification of Radical Expressions 1 Answer Don't Memorise Apr 4, 2016 #=28 sqrt2 # Explanation: #7 sqrt 32# We first prime factorise #32# ( express a number as a product of its prime factors) #sqrt 32 = sqrt ( 2 * 2 * 2 * 2 * 2 ) = sqrt (2^2 * 2^2 * 2) = 2 * 2 * sqrt2 = color(blue)(4sqrt2# #7 sqrt 32 = 7 * (color(blue)(4sqrt2))# #=28 sqrt2 # Answer link Related questions How do you simplify radical expressions? How do you simplify radical expressions with fractions? How do you simplify radical expressions with variables? What are radical expressions? How do you simplify #root{3}{-125}#? How do you write # ""^4sqrt(zw)# as a rational exponent? How do you simplify # ""^5sqrt(96)# How do you write # ""^9sqrt(y^3)# as a rational exponent? How do you simplify #sqrt(75a^12b^3c^5)#? How do you simplify #sqrt(50)-sqrt(2)#? See all questions in Simplification of Radical Expressions Impact of this question 1432 views around the world You can reuse this answer Creative Commons License